A car starting from rest moves on a straight path for time t = 0 to t = T with a uniform acceleration a and then stops with a uniform retardation. The average speed of the car will be :

1. $\dfrac{\text {aT}}{4}$

2. $\dfrac{3\text {aT}}{2}$

3. $\dfrac{\text {aT}}{2}$

4. aT

$\dfrac{\text {aT}}{2}$

Reason

Assuming, uniform retardation has the same value as the uniform acceleration (a).

Case 1 : When car is accelerating

Given,

Initial speed (u₁) = 0

Acceleration (a₁) = a

Time (t₁) = T

Let, distance travelled be S₁ and final velocity is v₁.

As,

v = u + at

On putting values

v₁ = u₁ + a₁t₁ = 0 + aT = aT

By using,

S = ut + $\dfrac{1}{2}$at²

On putting values

S₁ = u₁t₁ + $\dfrac{1}{2}$a₁$\text t_1^2$ = 0 + $\dfrac{1}{2}$aT² = $\dfrac{1}{2}$aT²

Case 2 : When car is retarding

Given,

Initial speed (u₂) = v₁ = aT

Acceleration (a₂) = -a

Final speed (v₂) = 0

Let, distance travelled be S₂ and time taken is t₂.

As,

v = u + at

On putting values

v₂ = u₂ + a₂t₂

⟹ 0 = aT - at₂

⟹ aT = at₂

⟹ t₂ = T

By using,

S = ut + $\dfrac{1}{2}$at²

On putting values

S₂ = u₂t₂ + $\dfrac{1}{2}$a₂$\text t_2^2$

= (aT)(T) - $\dfrac{1}{2}$aT²

= aT² - $\dfrac{1}{2}$aT²

= $\dfrac{1}{2}$aT²

Here,

Total distance travelled (S) = S₁ + S₂ = $\dfrac{1}{2}$aT² + $\dfrac{1}{2}$aT² = aT²

Total time taken (t) = t₁ + t₂ = T + T = 2T

Then,

$\text {Avg. Speed (v)}=\dfrac{\text {Total distance (S)}}{\text {Total time (t)}} \\[1em] =\dfrac{\text {aT}^2}{2\text T}= \dfrac{\text {aT}}{2}$

∴ Average speed of the car is $\dfrac{ {\bold {aT}}}{\bold 2}.$